If $r = 0$ then we find the equation gives $1=0$, so r cannot equal $0$.

Substituting $r^2$ for $a$, $-r$ for $b$ and $-(r-1)$ for $c$ in the usual quadratic equation formula we get
$$
X = \frac{r\pm\sqrt{r²+4r²(r-1)}}{2r^2}=\frac{1\pm\sqrt{1+4r-4}}{2r} = \frac{1\pm\sqrt{4r-3}}{2r}
$$
So this takes real values if and only if $4r - 3 \geq 0$.

Alex realised that the key feature of this problem was the discriminant of the quadratic equations in $X$, and looked at that object directly:

The discriminant determines whether a quadratic has real roots, therefore $X(r)$ will have real values if and only if the discriminant $D$ of the quadratic is zero or positive.

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